\(P=\dfrac{3+\sqrt{x}-3+\sqrt{x}}{9-x}\cdot\dfrac{3+\sqrt{x}}{\sqrt{x}}=\dfrac{2}{3-\sqrt{x}}\)
Để P>1/2 thì P-1/2>0
\(\Leftrightarrow-\dfrac{2}{\sqrt{x}-3}-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-3}+\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{4+\sqrt{x}-3}{2\left(\sqrt{x}-3\right)}< 0\)
=>\(\sqrt{x}-3< 0\)
=>0<x<9